Integrand size = 20, antiderivative size = 159 \[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\frac {\left (1+\frac {2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c \sqrt {x}}{b+\sqrt {b^2-4 a c}}\right )^{-p} (d x)^{1+m} \left (a+b \sqrt {x}+c x\right )^p \operatorname {AppellF1}\left (2 (1+m),-p,-p,3+2 m,-\frac {2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}},-\frac {2 c \sqrt {x}}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)} \] Output:
(d*x)^(1+m)*(a+b*x^(1/2)+c*x)^p*AppellF1(2+2*m,-p,-p,3+2*m,-2*c*x^(1/2)/(b -(-4*a*c+b^2)^(1/2)),-2*c*x^(1/2)/(b+(-4*a*c+b^2)^(1/2)))/d/(1+m)/((1+2*c* x^(1/2)/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^(1/2)/(b+(-4*a*c+b^2)^(1/2))) ^p)
Time = 0.62 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.15 \[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\frac {\left (\frac {b-\sqrt {b^2-4 a c}+2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c \sqrt {x}}{b+\sqrt {b^2-4 a c}}\right )^{-p} x (d x)^m \left (a+b \sqrt {x}+c x\right )^p \operatorname {AppellF1}\left (2 (1+m),-p,-p,3+2 m,-\frac {2 c \sqrt {x}}{b+\sqrt {b^2-4 a c}},\frac {2 c \sqrt {x}}{-b+\sqrt {b^2-4 a c}}\right )}{1+m} \] Input:
Integrate[(d*x)^m*(a + b*Sqrt[x] + c*x)^p,x]
Output:
(x*(d*x)^m*(a + b*Sqrt[x] + c*x)^p*AppellF1[2*(1 + m), -p, -p, 3 + 2*m, (- 2*c*Sqrt[x])/(b + Sqrt[b^2 - 4*a*c]), (2*c*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c ])])/((1 + m)*((b - Sqrt[b^2 - 4*a*c] + 2*c*Sqrt[x])/(b - Sqrt[b^2 - 4*a*c ]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*Sqrt[x])/(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1716, 1715, 1179, 150}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx\) |
\(\Big \downarrow \) 1716 |
\(\displaystyle x^{-m} (d x)^m \int x^m \left (a+c x+b \sqrt {x}\right )^pdx\) |
\(\Big \downarrow \) 1715 |
\(\displaystyle 2 x^{-m} (d x)^m \int x^{\frac {1}{2} (2 m+1)} \left (a+c x+b \sqrt {x}\right )^pd\sqrt {x}\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle 2 x^{-m} (d x)^m \left (\frac {2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c \sqrt {x}}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b \sqrt {x}+c x\right )^p \int \left (\frac {2 \sqrt {x} c}{b-\sqrt {b^2-4 a c}}+1\right )^p \left (\frac {2 \sqrt {x} c}{b+\sqrt {b^2-4 a c}}+1\right )^p x^{\frac {1}{2} (2 m+1)}d\sqrt {x}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {x (d x)^m \left (\frac {2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c \sqrt {x}}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b \sqrt {x}+c x\right )^p \operatorname {AppellF1}\left (2 (m+1),-p,-p,2 m+3,-\frac {2 c \sqrt {x}}{b-\sqrt {b^2-4 a c}},-\frac {2 c \sqrt {x}}{b+\sqrt {b^2-4 a c}}\right )}{m+1}\) |
Input:
Int[(d*x)^m*(a + b*Sqrt[x] + c*x)^p,x]
Output:
(x*(d*x)^m*(a + b*Sqrt[x] + c*x)^p*AppellF1[2*(1 + m), -p, -p, 3 + 2*m, (- 2*c*Sqrt[x])/(b - Sqrt[b^2 - 4*a*c]), (-2*c*Sqrt[x])/(b + Sqrt[b^2 - 4*a*c ])])/((1 + m)*(1 + (2*c*Sqrt[x])/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*Sqrt [x])/(b + Sqrt[b^2 - 4*a*c]))^p)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) ^p) Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m , p}, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k*(m + 1) - 1)*(a + b* x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, m, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && FractionQ[n]
Int[((d_)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_S ymbol] :> Simp[d^IntPart[m]*((d*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && EqQ[n2, 2 *n] && NeQ[b^2 - 4*a*c, 0] && FractionQ[n]
\[\int \left (d x \right )^{m} \left (a +b \sqrt {x}+c x \right )^{p}d x\]
Input:
int((d*x)^m*(a+b*x^(1/2)+c*x)^p,x)
Output:
int((d*x)^m*(a+b*x^(1/2)+c*x)^p,x)
Exception generated. \[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\text {Exception raised: TypeError} \] Input:
integrate((d*x)^m*(a+b*x^(1/2)+c*x)^p,x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: algl ogextint: unimplemented
Timed out. \[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\text {Timed out} \] Input:
integrate((d*x)**m*(a+b*x**(1/2)+c*x)**p,x)
Output:
Timed out
\[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\int { {\left (c x + b \sqrt {x} + a\right )}^{p} \left (d x\right )^{m} \,d x } \] Input:
integrate((d*x)^m*(a+b*x^(1/2)+c*x)^p,x, algorithm="maxima")
Output:
integrate((c*x + b*sqrt(x) + a)^p*(d*x)^m, x)
\[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\int { {\left (c x + b \sqrt {x} + a\right )}^{p} \left (d x\right )^{m} \,d x } \] Input:
integrate((d*x)^m*(a+b*x^(1/2)+c*x)^p,x, algorithm="giac")
Output:
integrate((c*x + b*sqrt(x) + a)^p*(d*x)^m, x)
Timed out. \[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\int {\left (d\,x\right )}^m\,{\left (a+c\,x+b\,\sqrt {x}\right )}^p \,d x \] Input:
int((d*x)^m*(a + c*x + b*x^(1/2))^p,x)
Output:
int((d*x)^m*(a + c*x + b*x^(1/2))^p, x)
\[ \int (d x)^m \left (a+b \sqrt {x}+c x\right )^p \, dx=\text {too large to display} \] Input:
int((d*x)^m*(a+b*x^(1/2)+c*x)^p,x)
Output:
(d**m*(4*x**((2*m + 1)/2)*(sqrt(x)*b + a + c*x)**p*b*m*p + 2*x**((2*m + 1) /2)*(sqrt(x)*b + a + c*x)**p*b*p**2 - 4*x**m*(sqrt(x)*b + a + c*x)**p*a*m* p - 2*x**m*(sqrt(x)*b + a + c*x)**p*a*p + 8*x**m*(sqrt(x)*b + a + c*x)**p* c*m**2*x + 12*x**m*(sqrt(x)*b + a + c*x)**p*c*m*p*x + 4*x**m*(sqrt(x)*b + a + c*x)**p*c*m*x + 4*x**m*(sqrt(x)*b + a + c*x)**p*c*p**2*x + 2*x**m*(sqr t(x)*b + a + c*x)**p*c*p*x - 16*int((x**((2*m + 1)/2)*(sqrt(x)*b + a + c*x )**p)/(4*a**2*m**3*x + 10*a**2*m**2*p*x + 6*a**2*m**2*x + 8*a**2*m*p**2*x + 9*a**2*m*p*x + 2*a**2*m*x + 2*a**2*p**3*x + 3*a**2*p**2*x + a**2*p*x + 8 *a*c*m**3*x**2 + 20*a*c*m**2*p*x**2 + 12*a*c*m**2*x**2 + 16*a*c*m*p**2*x** 2 + 18*a*c*m*p*x**2 + 4*a*c*m*x**2 + 4*a*c*p**3*x**2 + 6*a*c*p**2*x**2 + 2 *a*c*p*x**2 - 4*b**2*m**3*x**2 - 10*b**2*m**2*p*x**2 - 6*b**2*m**2*x**2 - 8*b**2*m*p**2*x**2 - 9*b**2*m*p*x**2 - 2*b**2*m*x**2 - 2*b**2*p**3*x**2 - 3*b**2*p**2*x**2 - b**2*p*x**2 + 4*c**2*m**3*x**3 + 10*c**2*m**2*p*x**3 + 6*c**2*m**2*x**3 + 8*c**2*m*p**2*x**3 + 9*c**2*m*p*x**3 + 2*c**2*m*x**3 + 2*c**2*p**3*x**3 + 3*c**2*p**2*x**3 + c**2*p*x**3),x)*a**2*b*m**5*p - 40*i nt((x**((2*m + 1)/2)*(sqrt(x)*b + a + c*x)**p)/(4*a**2*m**3*x + 10*a**2*m* *2*p*x + 6*a**2*m**2*x + 8*a**2*m*p**2*x + 9*a**2*m*p*x + 2*a**2*m*x + 2*a **2*p**3*x + 3*a**2*p**2*x + a**2*p*x + 8*a*c*m**3*x**2 + 20*a*c*m**2*p*x* *2 + 12*a*c*m**2*x**2 + 16*a*c*m*p**2*x**2 + 18*a*c*m*p*x**2 + 4*a*c*m*x** 2 + 4*a*c*p**3*x**2 + 6*a*c*p**2*x**2 + 2*a*c*p*x**2 - 4*b**2*m**3*x**2...